ZOJ 3203 Light Bulb （三分查找）

Light Bulb

---

Time Limit: 1 Second      Memory Limit: 32768 KB

---

Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is wandering in his incommodious house, thinking of how to earn more money. One day, he found that the length of his shadow was changing from time to time while walking between the light bulb and the wall of his house. A sudden thought ran through his mind and he wanted to know the maximum length of his shadow.

Input

The first line of the input contains an integer T (T <= 100), indicating the number of cases.

Each test case contains three real numbers H, h and D in one line. H is the height of the light bulb while h is the height of mildleopard. D is distance between the light bulb and the wall. All numbers are in range from 10^-2 to 10³, both inclusive, and H - h >= 10^-2.

Output

For each test case, output the maximum length of mildleopard's shadow in one line, accurate up to three decimal places..

Sample Input

3
2 1 0.5
2 0.5 3
4 3 4

Sample Output

1.000
0.750
4.000

---

Author: GUAN, Yao
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

二分法作为分治中最常见的方法，适用于单调函数，逼近求解某点的值。但当函数是凸性函数时，二分法就无法适用，这时三分法就可以“大显身手”~

 

如图，类似二分的定义Left和Right，mid = (Left + Right) / 2，midmid = (mid + Right) / 2;

如果mid靠近极值点，则Right = midmid；否则(即midmid靠近极值点)，则Left = mid;

模版如下：

int Cale(int  ){

}

int Solve(int left ,int right){
    int mid,midmid;
    while(left<right){
        mid=(left+right)/2;
        midmid=(mid+right)/2;
        if(Cale(mid) > Cale(midmid)) 
            right=midmid;///////假设求最大值
        else 
            left=mid;
    }
    return left;
}

如图，人左右走动，求影子L的最长长度。根据图，很容易发现当灯，人的头部和墙角成一条直线时(假设此时人站在A点)，此时的长度是影子全在地上的最长长度。当人再向右走时，影子开始投影到墙上，当人贴着墙，影子长度即为人的高度。所以当人从A点走到墙，函数是先递增再递减，为凸性函数，所以我们可以用  三分法 来求解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

const double eps=1e-8;

double H,h,D,ans;

double Cal(double l){
    return l+D*(h-l)/(H-l);
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%lf%lf%lf",&H,&h,&D);
        double left=0,right=h,mid,midmid;
        while(right-left>=eps){
            mid=(left+right)/2;
            midmid=(mid+right)/2;
            if(Cal(mid)>Cal(midmid))
                right=midmid;
            else
                left=mid;
        }
        printf("%.3lf
",Cal(left));
    }
    return 0;
}